A) 0.32 amp
B) 0.016 amp
C) 0.48 amp
D) 0.80 amp
Correct Answer: B
Solution :
\[i=\frac{V}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}=\frac{120}{\sqrt{100+4{{\pi }^{2}}\times {{60}^{2}}\times {{20}^{2}}}}\]=0.016 A
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